Question #12448

Use L'Hopital's rule to find limit x(pi/2 -tan^-1x) as x approaches infinity

Expert's answer

limit x(pi/2 -tan^-1x)_{x approaches infinity}=limit (pi/2 -tan^-1x)/(1/x)_{x approaches infinity}Both functions 1/x and_{ }(pi/2 -tan^-1x) tend to 0 when x tends to infinity/ So we use L'Hopital's rule. (1/x)'=-1/x^2

(pi/2 -tan^-1x)'=-1/(1+x^2)

limit (pi/2 -tan^-1x)/(1/x)_{x approaches infinity}=limit {-1/(1+x^2)/(-1/x^2)}_{x approaches infinity}=

=limit {x^2/(1+x^2)}_{x approaches infinity}=1

To know more about L'Hopital's rule you can watch our video which contains examples and detailed explanations. Take a look!

(pi/2 -tan^-1x)'=-1/(1+x^2)

limit (pi/2 -tan^-1x)/(1/x)

=limit {x^2/(1+x^2)}

To know more about L'Hopital's rule you can watch our video which contains examples and detailed explanations. Take a look!

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